Ta có : \(\begin{cases}8a+3⋮d\\5a+2⋮d\end{cases}\) \(\Leftrightarrow\begin{cases}5\left(8a+3\right)⋮d\\8\left(5a+2\right)⋮d\end{cases}\) \(\Leftrightarrow\begin{cases}40a+15⋮d\\40a+16⋮d\end{cases}\) \(\Rightarrow\left(40a+16\right)-\left(40a+15\right)⋮d\Rightarrow1⋮d\Rightarrow d\le1\)
Mà \(d\ge1\) \(\Rightarrow d=1\RightarrowƯCLN\left(8a+3,5a+2\right)=1\)
